Problem: Complete the square to solve for $x$. $x^{2}-12x+32 = 0$
Begin by moving the constant term to the right side of the equation. $x^2 - 12x = -32$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-12$ , half of it would be $-6$ , and squaring it gives us ${36}$ $x^2 - 12x { + 36} = -32 { + 36}$ We can now rewrite the left side of the equation as a squared term. $( x - 6 )^2 = 4$ Take the square root of both sides. $x - 6 = \pm2$ Isolate $x$ to find the solution(s). $x = 6\pm2$ So the solutions are: $x = 8 \text{ or } x = 4$ We already found the completed square: $( x - 6 )^2 = 4$